Answer:
8.85% per year
Step-by-step explanation:
To find the interest rate of a compounding interest, we use the formula:
![r=n[(\dfrac{A}{P})^{\dfrac{1}{nt}}-1]](https://tex.z-dn.net/?f=r%3Dn%5B%28%5Cdfrac%7BA%7D%7BP%7D%29%5E%7B%5Cdfrac%7B1%7D%7Bnt%7D%7D-1%5D)
Before we start solving, let's break down all the variables that we have.
A = 19,992.71
P = 10,000.00
n = 2
t = 8
r = ?
Now let's put the values into the formula.
![r=2[(\dfrac{19,992.71}{10,000.00})^{\dfrac{1}{2(8)}}-1]](https://tex.z-dn.net/?f=r%3D2%5B%28%5Cdfrac%7B19%2C992.71%7D%7B10%2C000.00%7D%29%5E%7B%5Cdfrac%7B1%7D%7B2%288%29%7D%7D-1%5D)
![r=2[(\dfrac{19,992.71}{10,000.00})^{\dfrac{1}{16}}-1]](https://tex.z-dn.net/?f=r%3D2%5B%28%5Cdfrac%7B19%2C992.71%7D%7B10%2C000.00%7D%29%5E%7B%5Cdfrac%7B1%7D%7B16%7D%7D-1%5D)
![r=2[(\dfrac{19,992.71}{10,000.00})^{0.0625}-1]](https://tex.z-dn.net/?f=r%3D2%5B%28%5Cdfrac%7B19%2C992.71%7D%7B10%2C000.00%7D%29%5E%7B0.0625%7D-1%5D)
![r=2[(1.9992.1)^{0.0625}-1]](https://tex.z-dn.net/?f=r%3D2%5B%281.9992.1%29%5E%7B0.0625%7D-1%5D)
![r=2[1.0442499885-1]](https://tex.z-dn.net/?f=r%3D2%5B1.0442499885-1%5D)
![r=2[0.0442499885]](https://tex.z-dn.net/?f=r%3D2%5B0.0442499885%5D)
or 
So the rate the Mrs. Emily Francis got from the bank was 8.85% per year.
Answer:
Step-by-step explanation:
If θ is the angle between two vectors u and v, then cosθ = (u·v) / [llull llvll]
u·v = (8)(9) + (7)(7) = 121
llull = √[(8)2 + (9)2] = √145 llvll = √[(7)2+(7)2] = 7√2
So, cosθ = 121/ [7√290]
= 1.015052094
θ = Cos-1(1.015052094) ≈ 3.3
Step-by-step explanation:
=11/3×1/14
=3.67×0.07
=0.2568
Answer:
10 pounds of dog food.
Step-by-step explanation:
You can find this answer by multiplying 1/4 to 40
Because
There is 40 small dogs and each eats 1/4 of dog food
Answer:

Step-by-step explanation:
Previous concepts
Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".
The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".
Solution to the problem
Let X the random variable that represent the IQ scores of a population, and for this case we know the distribution for X is given by:
Where
and
If we standardize the variable with the z score given by:

We got a normal standard distribution with parameters 
